Question

A 78.0 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.190 m/s. How much work must be done on the hoop to stop it?

Answer 1

Givens: v_cm-i = 0.190 m/s and v_cm-f = 0 and the mass m = 78.0 kg

As the hoop is rolling we have to take into account the physics of rigid body:

Moment of Inertia I = mr^2 and the respective rotational kinetic energy of the hoop: KE_rotational = Iw^2 / 2

The total Inetial kinetic energy is equal to the sum of the rotational and linear kinetic energy:

KE_^total = KE_linear + KE_rotational

_{plugging the expression of the moment of inertia of the hoop, we get:}

and we can use the relation between the angular velocity and the velocity of the center of mass: v_cm = r*w

Now we can use the expression that defined the wrok done to stop it:

the total kinetic energy is equal to zero, as the hoop stopped: KE_total-final = 0 and plugging it into our last equation we get the value of the work:

and the work necessary to stop the hoop is 2.8158 J (the negative sign we got, is because the force to stop it must be against the motion):